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aivan3 [116]
2 years ago
5

Part 2: a) Write VHDL code for a top module that invokes necessary components to display the four decimal digits on four seven-s

egment displays of a NEXYX 4 FPGA. The top module takes the 100 MHz FPGA clock, a reset signal, an enabler signal as the inputs. b) Makes necessary changes in a constraint file to implement your design on the FPGA and present only the uncommented parts of the constraint file in your exam script.
Computers and Technology
1 answer:
dimulka [17.4K]2 years ago
7 0
I don’t know ‍♀️ sorry
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#A year is considered a leap year if it abides by the #following rules: # # - Every 4th year IS a leap year, EXCEPT... # - Every
lara [203]

Answer:

To check if the year comes under each 100th year, lets check if the remainder when dividing with 100 is 0 or not.

Similarly check for 400th year and multiple 0f 4. The following C program describes the function.

#include<stdio.h>

#include<stdbool.h>

bool is_leap_year(int year);

void main()

{

int y;

bool b;

 

printf("Enter the year in yyyy format: e.g. 1999 \n");

scanf("%d", &y);     // taking the input year in yyyy format.

 

b= is_leap_year(y);  //calling the function and returning the output to b

if(b==true)

{

 printf("Thae given year is a leap year \n");

}

else

{

 printf("The given year is not a leap year \n");

}

}

bool is_leap_year(int year)

{

if(year%100==0)   //every 100th year

{

 if(year%400==0)   //every 400th year

 {

  return true;

 }

 else

 {

  return false;

 }

}

if(year%4==0)  //is a multiple of 4

{

 return true;

}

else

{

 return false;

}

}

Explanation:

Output is given as image

5 0
3 years ago
How do users log into Windows 8?
bogdanovich [222]
I think it’s swipe up if i’m correct, if not sorry
5 0
3 years ago
Yo, how can I bypass the securely filter on a chromebook?
OLga [1]
Open your Chromebook and press the power button for 30 seconds. This should bypass the admin block.

Proxy Websites. Probably one of the oldest methods of bypassing web filters, proxy websites enable users to anonymously connect to websites through outside servers. ...
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5 0
3 years ago
Write a function listLengthOfAllWords which takes in an array of words (strings), and returns an array of numbers representing t
vesna_86 [32]

Answer:

   public static int[] listLengthOfAllWords(String [] wordArray){

       int[] intArray = new int[wordArray.length];

       for (int i=0; i<intArray.length; i++){

           int lenOfWord = wordArray[i].length();

           intArray[i]=lenOfWord;

       }

       return intArray;

   }

Explanation:

  1. Declare the method to return an array of ints and accept an array of string as a parameter
  2. within the method declare an array of integers with same length as the string array received as a parameter.
  3. Iterate using for loop over the array of string and extract the length of each word using this statement  int lenOfWord = wordArray[i].length();
  4. Assign the length of each word in the String array to the new Integer array with this statement intArray[i]=lenOfWord;
  5. Return the Integer Array

A Complete Java program with a call to the method is given below

<em>import java.util.Arrays;</em>

<em>import java.util.Scanner;</em>

<em>public class ANot {</em>

<em>    public static void main(String[] args) {</em>

<em>       String []wordArray = {"John", "James", "David", "Peter", "Davidson"};</em>

<em>        System.out.println(Arrays.toString(listLengthOfAllWords(wordArray)));</em>

<em>        }</em>

<em>    public static int[] listLengthOfAllWords(String [] wordArray){</em>

<em>        int[] intArray = new int[wordArray.length];</em>

<em>        for (int i=0; i<wordArray.length; i++){</em>

<em>            int lenOfWord = wordArray[i].length();</em>

<em>            intArray[i]=lenOfWord;</em>

<em>        }</em>

<em>        return intArray;</em>

<em>    }</em>

<em>}</em>

This program gives the following array as output: [4, 5, 5, 5, 8]

7 0
3 years ago
What is booting as used in computers?​
Sliva [168]

Answer:

the startup process

Explanation:

5 0
3 years ago
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